Well, it's not so obvious that under any circumstances all macroscopic systems behave classically, and indeed there are examples, where this is not the case, e.g., superfluidity, BECs, superconductivity, etc. are examples where you have specific quantum behavior of macroscopic systems.RockyMarciano said:But isn't standard QT(not interpretational) that in the macroscopic limit the expectation value and the Born rule are equivalent, and one can use either one to derive the other, and therefore eithe one can be used as stariting postulate?
This is my point in that post. Regardless of the shortcomings of the Born rule, I don't think Neumaier's foundational attack to Born's rule from within the theory is possible.vanhees71 said:Yes, but they talk in the usual probabilistic language, and I've given this explanation for the classicality of many macroscopic observables in the very beginning of this thread. The short paper is a very nice demonstration of how QT is consistent in describing the object and measurement apparatus, which is macroscopic, quantum mechanically. It does not deny or disprove standard quantum theory and/or Born's rule!
They are not equivalent.RockyMarciano said:But isn't standard QT(not interpretational) that in the macroscopic limit the expectation value and the Born rule are equivalent, and one can use either one to derive the other, and therefore either one can be used as starting postulate?
Only because it is difficult to avoid using this language. But they don't regard Born's rule as part of the foundation of quantum theory.vanhees71 said:but they talk in the usual probabilistic language,
I don't know why you insist on this. It is a basic assumption in QM and statisitical mechanics that macroscopic objects are made of large numbers of microscopic objects and that according to this single measurements done to macroscopic objects are FAPP equivalent to many repeated measurements of those microscopic constituents and in this sense to their expectation value when its uncertainty tends to zero.A. Neumaier said:They are not equivalent.
One can never derive from a rule that only applies to many repeated measurements anything that applies to a single measurement.
No. They aren't. A measurement is something that is actually measured and yields a communicatable numerical result.RockyMarciano said:according to this single measurements done to macroscopic objects are FAPP equivalent to many repeated measurements of those microscopic constituents
This distinction is irrelevant in the context of the QT formalism and statisitical mechanics, which you claim you are not questioning. Every single textbook on QT assumes that lab single measurements results in practice are useful precisely because one can imply that the single macroscopic measurement is a simultaneous measure of very many microscopic constituents in the same state, and that this is also equivalent to measuring repeatedly a single system.A. Neumaier said:No. They aren't. A measurement is something that is actually measured and yields a communicatable numerical result.
Thus many repeated experiments yield many recorded results. But a single macroscopic measurement yields only a single recorded result.
How does the single number measure many things simultaneously?RockyMarciano said:the single macroscopic measurement is a simultaneous measure of very many microscopic constituents in the same state,
Please tell me the measurement procedure used to measure each of these many microscopic constituents, and the measurement system with which these astronomically many measurements are carried out.ISO said:2.1
measurement
process of experimentally obtaining information about the magnitude of a quantity
NOTES
1 Measurement implies a measurement procedure, based on a theoretical model.
2 In practice, measurement presupposes a calibrated measuring system, possibly
subsequently verified.
It boils down to the point that theoretical expectation values and means over frequently repeated measurements are not the same thing, and that the former makes sense even for cases where no repetitions are made. But this is a hollow phrase only if one doesn't realize the implications of it for the measurement problem.mikeyork said:Although I haven't attempted to follow all the details of the discussion here, I can't help feeling that it boils down to the issue I raised in post #3: that theoretical probability and statistical relative frequency are not necessarily the same thing.
Well, a theoretical expectation value is a property of a distribution function is it not? So the issue becomes one of interpretation of the distribution.A. Neumaier said:It boils down to the point that theoretical expectation values and means over frequently repeated measurements are not the same thing, and that the former makes sense even for cases where no repetitions are made. But this is a hollow phrase only if one doesn't realize the implications of it for the measurement problem.
No. Only for randon variables. But quantum operators are not random variables. There is no a priori reason why ##Tr\rho A## should have anything to do with measurement at all. Thus the connection, if any, must be postulated by an interpretation.mikeyork said:a theoretical expectation value is a property of a distribution function is it not?
So why call it an expectation value? AFAIK, the term is used exclusively for the first moment of a distribution function. Even a deterministic variable has a distribution function -- it's a delta-function.A. Neumaier said:No. Only for randon variables
They are not any sort of variable. They are operators.A. Neumaier said:But quantum operators are not random variables.
Yes, and They behave like random variables only under special circumstances, and like deterministic variables even more rarely.mikeyork said:They are not any sort of variable. They are operators.
By analogy. Things are often called by names borrowed from a special case or an analogy.mikeyork said:So why call it an expectation value?
But an operator is a priori a mathematical object and not a measurable variable and hence has no distribution. Thus its expectation value is a purely theoretical quantity unless you declare by an interpretation rule how it is related to measurement. The interpretation rule used by the authors of the paper under discussion is the one given by the thermal interpretation, which makes no reference to a distribution.mikeyork said:Every measurable variable has a distribution -
I never claimed otherwise. I said any variable has a distribution function. That distribution function is what the Born rule provides.A. Neumaier said:But an operator is a priori a mathematical object and not a measurable variable and hence has no distribution. Thus its expectation value is a purely theoretical quantity unless you declare by an interpretation rule how it is related to measurement.
Here is where you have not grasped my point. A distribution function is a theoretical quantity derived from a theoretical concept of probability that applies directly to any single measurement regardless of whether you might (or could) count frequencies.No distributions are needed (or even sensible) to interpret a single measurement. The measurement gives a result, and that's it. Distributions make sense only if you can repeat the measurement arbitrarily often.
But an operator is not a variable.mikeyork said:I said any variable has a distribution function.
Does a measurement measure an operator or a variable? Do we measure ##<A>## or ##<a>##.A. Neumaier said:But an operator is not a variable.
The expectation ##\langle a\rangle## is a variable, not an operator and the variable ##a##, with values given by the eigenvalues of ##A##, has a distribution function ##|<a|\psi>|^2## for every state ##\psi##. That is the entire content of Born rule. It says nothing about operators; only variables. ##<a|\psi>## is what matters not ##A##.A non-normal operator such as ##A=p+iq## has not even in principle a distribution function (since measuring ##A## would mean measuring ##p## and ##q##), though its expectation ##\langle A\rangle:=Tr \rho A## is well-defined for every state ##\rho##!
What are you referring to? We measure the expectation value of the observable variable ##a## (a scalar) not the operator ##A##. An operator means nothing until you operate on something.Jilang said:Sorry, perhaps I am missing something, but it seems to me that <A> says as much about A as it does about the density function, given they both appear on the RHS.
What is measured depends on the kind of measurement.mikeyork said:Does a measurement measure an operator or a variable? Do we measure ##<A>## or ##<a>##.
No, it is a number.mikeyork said:The expectation ##\langle a\rangle## is a variable
What kind of macroscopic measurement do you have in mind? If we measure anything we get a number ##a_{macro}##. I have no idea what ##<A>## means except as shorthand for ##<\psi|A|\psi>## which is ##<a>##. You wrote ##<A> = Tr \rho A ## but ##Tr \rho A = <a>##. As far as I can see, ##<A>## (the "expectation value of an operator") may be a useful mathematical idea but is physically meaningless until you operate on something.A. Neumaier said:In a macroscopic measurement we measure ##<A>##
Ok, but my point is that we don't measure ##<A>##. And whatever variable we do measure, it has a prior (i.e. conditional on state preparation) distribution function ##|<a|\psi>|^2##.in a Born-type measurement we measure an eigenvalue of ##A##